For Numpy and Tensor,
For List, cannot put a list into a list
libstdc++.so.6: version `GLIBCXX_3.4.29' not found
If has error when you install pycocotools, use:
For static batchsize:
/usr/src/tensorrt/bin/trtexec --onnx=model.onnx --saveEngine=model.engine --explicitBatch --fp16 --verbose --workspace=8192 --dumpOutput
For dynamic batchsize:
/usr/src/tensorrt/bin/trtexec --onnx=model.onnx --saveEngine=model.engine --explicitBatch --fp16 --verbose --dumpOutput --minShapes='input':1x3x640x640 --optShapes='input':6x3x640x640 --maxShapes='input':12x3x640x640 --exportTimes=trace.json --dumpProfile --exportProfile=prof.json
Code:
import os
for (root, dirs, file) in os.walk(path):
for f in file:
if '.txt' in f:
print(f)
Code:
import matplotlib.path as mplPath
def check_in_roi(bbox, roi, rw, rh):
x1 = bbox[0]
y1 = bbox[1]
x2 = bbox[2]
y2 = bbox[3]
feet_dot_x = int((x1+x2)/2)
feet_dot_y = y2
temp = []
for c in roi:
temp.append([c[0]*rw, c[1]*rh])
roi_path = mplPath.Path(np.array(temp))
if roi_path.contains_point((feet_dot_x, feet_dot_y)):
return True
else:
return Falseinput roi should be a list of dot [x,y]
Python Code for IOU (numpy):
def iou_nms(bbox,gt):
#lt是两个框中间重叠框的最左边和最上边的坐标，
#rb是两个框中间重叠框的最右边和最下边的坐标
lt = np.maximum(bbox[:,None,:2],gt[:,:2]) # [N,M,2]
rb = np.minimum(bbox[:,None,2:4],gt[:,2:4]) # [N,M,2]
#wh是重叠框的宽和高，+1是因为求边长，边长就等于前后两端的坐标点相减且+1
wh = np.maximum(rb - lt + 1 , 0) # [N,M,2]
#求重叠框的面积
inter_area = wh[:,:,0] * wh[:,:,1] #[N,M]
#分别求两个框各自的面积
bbox_area = (bbox[:,2] - bbox[:,0] + 1) * (bbox[:,3] - bbox[:,1] + 1) #[N,]
gt_area = (gt[:,2] - gt[:,0] + 1) * (gt[:,3] - gt[:,1] + 1) #[M,]
#iou的公式，重叠框面积 / 两个框面积之和减去重叠框面积
iou = inter_area / (bbox_area[:,None] + gt_area - inter_area) #[N,M]
return iouinput: bbox and gt should be numpy
Python Code for IOU (list):
def iou(boxA, boxB): # determine the (x, y)-coordinates of the intersection rectangle xA = max(boxA[0], boxB[0]) yA = max(boxA[1], boxB[1]) xB = min(boxA[2], boxB[2]) yB = min(boxA[3], boxB[3]) # compute the area of intersection rectangle interArea = max(0, xB - xA + 1) * max(0, yB - yA + 1) # compute the area of both the prediction and ground-truth # rectangles boxAArea = (boxA[2] - boxA[0] + 1) * (boxA[3] - boxA[1] + 1) boxBArea = (boxB[2] - boxB[0] + 1) * (boxB[3] - boxB[1] + 1) # compute the intersection over union by taking the intersection # area and dividing it by the sum of prediction + ground-truth # areas - the interesection area iou = interArea / float(boxAArea + boxBArea - interArea) # return the intersection over union value return iou
input: boxA and boxB should be list (format xyxy)
Python Code for NMS:
def nms(bbox,thresh):
#得分bbox第五列是得分，前四列是x0,y0,x1,y1
score = bbox[:,4]
#对得分进行排序
order = np.argsort(score)
#记录结果值，每次保存得分最高的那个框的索引，最后再用bbox[keep]取出相应框
keep = []
#一直筛选到没有可用的框
while order.size > 0:
#取得分最高的框的索引，因为order是升序，所以最后一位是得分最高的
index = order[-1]
#保存得分最高的那个框的索引
keep.append(index)
#取出这个框
x = bbox[index]
#计算iou，x[None,:]是为了保持shape一致，squeeze(0)是去掉第一个维度，
#不去掉的话结果的shape是[1,5]，再np.where就不对了，必须让其等于[5,]
sub_bbox_iou = iou_nms(x[None,:],bbox[order[:-1]]).squeeze(0)
#筛选小于阈值的框，大于阈值的话，就和得分最高的那个框重叠了，所以保留不重叠的框
index_after = np.where(sub_bbox_iou < thresh)
#筛选剩下的框
order = order[index_after]
return keep
* Input bbox should be numpy
* should be used with function iou_nms
* output is a list of index of the keep bbox
